While golfing, Sahar hits a tee shot from T toward a hole at H, but the ball veers 23 degrees and lands at B. The scorecard says that H is 270 m from T. If Sahar walks 160 m to the ball(B), how far, to the nearest metre, is the ball from the hole?
Did you make your sketch??
According to mine it is a simple case of the cosine law
HB^2 = 270^2 + 160^2 - 2(270)(160)cos 23°
= ..
take over
To find the distance from the ball (B) to the hole (H), we can use the concept of trigonometry. Since the ball veered 23 degrees off the desired path, we can consider it as the angle between the line from T to B and the line from T to H.
First, let's find the horizontal distance between the ball (B) and the hole (H). We can use the cosine function to calculate this distance:
Cos(angle) = Adjacent / Hypotenuse
In this case, the adjacent side is the horizontal distance between B and H, which is the distance we're trying to find. The hypotenuse is the distance from T to H, given as 270 m in the problem.
Cos(23 degrees) = Adjacent / 270 m
Now, we can rearrange the equation to solve for the adjacent side (distance between B and H):
Adjacent = Cos(23 degrees) * 270 m
Using a scientific calculator, the value of Cos(23 degrees) is approximately 0.9205.
Adjacent = 0.9205 * 270 m
Adjacent ≈ 248.22 m (rounded to two decimal places)
Therefore, the horizontal distance between the ball and the hole is approximately 248.22 meters.
To find the total distance from B (where the ball landed) to H (the hole), we need to consider the vertical distance as well. Since there is no given information about any vertical displacement or elevation changes, we can assume the vertical distance is negligible, and only consider the horizontal distance.
Thus, the distance from the ball B to the hole H is approximately 248 meters (rounded to the nearest meter).